The determinant has the properties:
det ( A B ) = det ( A ) det ( B ) det ( A ) = det ( A T )
Hence, given that ∣ ∣ ∣ ∣ ∣ ∣ a 1 a 4 a 7 a 2 a 5 a 8 a 3 a 6 a 9 ∣ ∣ ∣ ∣ ∣ ∣ = 1 6
Find ∣ ∣ ∣ ∣ ∣ ∣ 4 a 1 2 a 4 4 a 7 2 a 2 a 5 2 a 8 4 a 3 2 a 6 4 a 9 ∣ ∣ ∣ ∣ ∣ ∣
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If we use permutations ("Sarrus' Rule") to find the determinant, we see that every term is multiplied by 16, 1 6 a 1 a 5 a 9 + . . . − 1 6 a 3 a 5 a 7 , so that the determinant we seek is 1 6 2 = 2 5 6 .
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Let us call the first matrix A and the second B .
We see we can get B from A by applying the following operations:
Where R n → k R n multiplies every element in row n by k , and Where C n → k C n multiplies every element in column n by k . These operations can be represented by the matrices:
E 1 = ⎝ ⎛ 2 0 0 0 1 0 0 0 1 ⎠ ⎞ and E 2 = ⎝ ⎛ 1 0 0 0 1 0 0 0 2 ⎠ ⎞ .
Where pre-multipying is a row operation and post-multiplying is a column operation.
Then B = E 1 E 2 A E 1 E 2 , so det ( B ) = det ( E 1 ) 2 det ( E 2 ) 2 det ( A ) = 2 2 ⋅ 2 2 ⋅ 1 6 = 2 5 6 .